In the two-body problem, other bound-state levels can be ignored because that energy difference between levels are large.  But how about the the mixture with the continuous state.  

A related question.  What happens for two channels coupled at the two continium? 


\emph{What exactly is the difference between the 1-D Fourier transformation vs. the r-coordinator of the 3-d transformation?}

It is different. For a function $f(\vec{x})=f(x)$, the 3-D Fourier tansformation is 
\begin{equation}
\begin{split}
f(\vp)&=f(p)\\
	&=2\sqrt{2\pi}\int^\infty_0dx\,f(x)\frac{x}{p}\sin{px}
\end{split}
\end{equation}



